5x^2+4x=0 eching

$$5 x ^ { 2 } + 4 x = 0$$

x uchun yechish

$$x=-\frac{4}{5}=-0,8$$

$$x=0$$

Faktorlash qadamlari

x omili.

$$x\left(5x+4\right)=0$$

Tenglamani yechish uchun x=0 va 5x+4=0 ni yeching.

$$x=0 x=-\frac{4}{5}$$

Kvadrat tenglamalaridan foydalanish qoidalari

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$5x^{2}+4x=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 5 ni a, 4 ni b va 0 ni c bilan almashtiring.

$$x=\frac{-4±\sqrt{4^{2}}}{2\times 5}$$

4^{2} ning kvadrat ildizini chiqarish.

$$x=\frac{-4±4}{2\times 5}$$

2 ni 5 marotabaga ko'paytirish.

$$x=\frac{-4±4}{10}$$

x=\frac{-4±4}{10} tenglamasini yeching, bunda ± musbat. -4 ni 4 ga qo'shish.

$$x=\frac{0}{10}$$

0 ni 10 ga bo'lish.

$$x=0$$

x=\frac{-4±4}{10} tenglamasini yeching, bunda ± manfiy. -4 dan 4 ni ayirish.

$$x=-\frac{8}{10}$$

\frac{-8}{10} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$x=-\frac{4}{5}$$

Tenglama yechildi.

$$x=0 x=-\frac{4}{5}$$

Kvadratni yakunlash bo'yicha qadamlar

Bu kabi kvadrat tenglamalarni kvadratni yakunlab yechish mumkin. Kvadratni yechish uchun tenglama avval ushbu shaklda bo'lishi shart: x^{2}+bx=c.

$$5x^{2}+4x=0$$

Ikki tarafini 5 ga bo‘ling.

$$\frac{5x^{2}+4x}{5}=\frac{0}{5}$$

5 ga bo'lish 5 ga ko'paytirishni bekor qiladi.

$$x^{2}+\frac{4}{5}x=\frac{0}{5}$$

0 ni 5 ga bo'lish.

$$x^{2}+\frac{4}{5}x=0$$

\frac{4}{5} ni bo‘lish, x shartining koeffitsienti, 2 ga \frac{2}{5} olish uchun. Keyin, \frac{2}{5} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$x^{2}+\frac{4}{5}x+\left(\frac{2}{5}\right)^{2}=\left(\frac{2}{5}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib \frac{2}{5} kvadratini chiqarish.

$$x^{2}+\frac{4}{5}x+\frac{4}{25}=\frac{4}{25}$$

x^{2}+\frac{4}{5}x+\frac{4}{25} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(x+\frac{2}{5}\right)^{2}=\frac{4}{25}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(x+\frac{2}{5}\right)^{2}}=\sqrt{\frac{4}{25}}$$

Qisqartirish.

$$x+\frac{2}{5}=\frac{2}{5} x+\frac{2}{5}=-\frac{2}{5}$$

Tenglamaning ikkala tarafidan \frac{2}{5} ni ayirish.

$$x=0 x=-\frac{4}{5}$$